Question: Divide the following rational expressions and simplify the result. $\dfrac{10n^2+23n-5}{4n^2+6n} \div \dfrac{25n^2-10n+1}{7n+3}=$
Explanation: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $10n^2+23n-5$, of the first expression can be factored by grouping to $(2n+5)(5n-1)$. The denominator, $4n^2+6n$, of the first expression can be factored as $2n(2n+3)$ by factoring out $2n$. The numerator, $25n^2-10n+1$, of the second expression can be factored as $(5n-1)(5n-1)$ using the perfect squares pattern. The denominator, $7n+3$, of the second expression cannot be factored further. Now the quotient looks as follows: $\dfrac{(2n+5)(5n-1)}{2n(2n+3)} \div \dfrac{(5n-1)(5n-1)}{7n+3}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{(2n+5)(5n-1)}{2n(2n+3)} \div \dfrac{(5n-1)(5n-1)}{7n+3}$ $\begin{aligned} &= \dfrac{(2n+5)(5n-1)}{2n(2n+3)} \!\cdot\! \dfrac{7n+3}{(5n-1)(5n-1)} &\text{Flip the divisor.}\\\\\\ &= \dfrac{(2n+5)(5n-1) \cdot (7n+3)}{2n(2n+3) \cdot (5n-1)(5n-1)} &\text{Multiply across.}\\\\\\ &= \dfrac{(2n+5){\cancel{(5n-1)}} (7n+3)}{2n(2n+3) {\cancel{(5n-1)}}(5n-1)} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(2n+5) (7n+3)}{2n(2n+3)(5n-1)} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{(2n+5) (7n+3)}{2n(2n+3)(5n-1)}$, which is equivalent to $\dfrac{14n^2+41n+15}{20n^3+26n^2-6n}$.